2013
DOI: 10.1002/jgt.21734
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On Toughness and Hamiltonicity of 2K2‐Free Graphs

Abstract: The toughness of a (noncomplete) graph G is the minimum value of t for which there is a vertex cut A whose removal yields |A|/t components. Determining toughness is an NP‐hard problem for general input graphs. The toughness conjecture of Chvátal, which states that there exists a constant t such that every graph on at least three vertices with toughness at least t is hamiltonian, is still open for general graphs. We extend some known toughness results for split graphs to the more general class of 2K2‐free graph… Show more

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Cited by 35 publications
(39 citation statements)
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“…By lemma 2.2 (3,4), S is an independent set. Let T = V (P ) − S. By Lemma 2.2 (1,2), V (P ) contains at least 2k + 1 vertices.…”
Section: To Prove Assertion (3) Assume By Contradiction Thatmentioning
confidence: 96%
“…By lemma 2.2 (3,4), S is an independent set. Let T = V (P ) − S. By Lemma 2.2 (1,2), V (P ) contains at least 2k + 1 vertices.…”
Section: To Prove Assertion (3) Assume By Contradiction Thatmentioning
confidence: 96%
“…Considerable work went into investigating this computational problem for various classes of graphs. In particular, recently, Broersma, Patel and Pyatkin proved [6] that toughness of a 2K 2 -free graph, i.e. a graph that does not contain an induced copy of the disjoint union of two edges, can be found in polynomial time.…”
Section: Introductionmentioning
confidence: 99%
“…figure on the right, where we have a 2K 2 -free graph with a 2-walk, but without Hamiltonian prism. It is worth mentioning that the toughness constant in Theorem 2 is much better than 3 2 , the lower bound on toughness of a 2K 2 -free graph needed for its Hamiltonicity, see [6,Sect. 4].…”
Section: Introductionmentioning
confidence: 99%
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“…They are hamiltonian (even if the sets I v have different cardinalities, as long as the graphs are 1-tough. We refer to [16], where these graphs are called C * 5 -type graphs and treated as special cases of triangle-free 2K 2free graphs). Using that C 5 is K 1 ∪ P 4 -free, it is easy to check that all these graphs are K 1 ∪ P 4 -free.…”
Section: Theorem 63 (Bedrossianmentioning
confidence: 99%